Let $F=\cos(y^{2}+z^{2})i+\sin(z^{2}+x^{2})j+e^{x^{2}+y^{2}}k$ be a vectore field on $\mathbb R^{3}$. Calculate $\int_{S}F.ds$, where the surface $S$ is defined by $x^{2}+y^{2}=e^{z}\cos z~$, $~~0\leq z \leq \pi/2$ and oriented upward.
Is my solution is correct?
Your sketch of the surface is not right, by the way. Since $\frac d{dz}(e^z\cos z) = e^z(\cos z-\sin z)$, the surface gets fatter until $z=\pi/4$, and then thinner all the way to a point at $z=\pi/2$. "Oriented upward" is probably not a good phrase. However, when you glue on the bottom disk, its outward-pointing normal needs to be $(0,0,-1)$. When you take the negative of this flux integral, you get the flux with the upward-pointing normal $(0,0,1)$, which you did compute. (There's no need to compute a cross-product here!!) As @PaulSinclair pointed out, your solution has cancelling negative signs but comes out correct.